1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Elden [556K]
2 years ago
7

A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract

ure, given that: the applied load is 5560 N the flexural strength is 105 MPa the separation between the supports is 45 mm Input your answer as X.XX mm, but without the unit of mm.
Engineering
1 answer:
Hitman42 [59]2 years ago
8 0

Answer:

radius = 9.1 × 10^{-3} m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support =  45 mm

solution

we apply here minimum radius formula that is

radius = \sqrt[3]{\frac{FL}{\sigma \pi}}      .................1

here F is applied load and  is length

put here value and we get

radius =  \sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}  

solve it we get

radius = 9.1 × 10^{-3} m

You might be interested in
The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin
IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

\dot{m}=\rho AV

Putting all the value to get the velocity of the flow

\frac{\dot{m}}{\rho A} = V

V = \frac{4000}{1000*4*2}

V = 0.5 m/s

4 0
3 years ago
A battery is an electromechanical device. a)- True b)- False
lilavasa [31]

Answer:

b)False

Explanation:

A battery is a device which store the energy in the form of chemical energy.And this stored energy is used according to the requirement.So battery is not a electromechanical device.Because it does have any mechanical component like gear ,shaft flywheel etc.

A flywheel is known as mechanical battery because it stored mechanical energy and supply that energy when more energy is required.Generally fly wheel is used during punching operation.

5 0
3 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
3 years ago
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra
gladu [14]

Answer:

a) T_{2}=837.2K

b) e=91.3 %

Explanation:

A) First, let's write the energy balance:

W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})  (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 \frac{kJ}{kgK} And its specific R constant is 0.287 \frac{kJ}{kgK}.

The only unknown from the energy balance is T_{2}, so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K

B) The isentropic efficiency (e) is defined as:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}

Where {h_{2s} is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because h_{2}-h_{1} can be obtained from the energy balance  \frac{W}{m}=h_{2}-h_{1}

h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}

An entropy change for an ideal gas with  constant Cp is given by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})

Applying logarithm properties:

ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}

Then,

T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K

So, now it is possible to calculate h_{2s}-h_{1}:

h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}

Finally, the efficiency can be calculated:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3 %

4 0
3 years ago
Identify which sound type each line contains.
nydimaria [60]

Answer:i can not see it

Explanation:

4 0
2 years ago
Other questions:
  • Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high
    5·1 answer
  • what should be used to feed material into a machine? A.joy stick B.push stick C. your feet D. your hands​
    14·1 answer
  • What are some possible reasons for the sudden development of the cell theory
    14·1 answer
  • Which energy source would you rank as the best option for the rescue team? Why?
    9·1 answer
  • Independent auto lots usually have ____ finance rates than dealerships
    7·1 answer
  • A 200-gr (7000 gr = 1 lb) bullet goes from rest to 3300 ft/s in 0.0011 s. Determine the magnitude of the impulse imparted to the
    10·1 answer
  • Please choose a specific type of stability or control surface (e.g., a canard) and explain how it is used, what it is used for,
    5·1 answer
  • Sarah and Raj take/takes me to a baseball game every year.
    11·1 answer
  • Engineered lumber should not be used for
    15·1 answer
  • Why are plastics known as synthetic materials?​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!